Prove the formula that: ((∃x)(F(x)∧S(x))→(∀y)(M(y)→W(y)))∧((∃y)(M(y)∧¬W(y))) ⇒(∀x)(F(x)→¬S(x))

Question

Prove the formula that:
((∃x)(F(x)∧S(x))→(∀y)(M(y)→W(y)))∧((∃y)(M(y)∧¬W(y)))
⇒(∀x)(F(x)→¬S(x))

in progress 0
Nguyệt Ánh 6 months 2021-08-24T04:05:01+00:00 1 Answers 2 views 0

Answers ( )

    0
    2021-08-24T04:06:18+00:00

    Step-by-step explanation:

    Given: [∀x(L(x) → A(x))] →

    [∀x(L(x) ∧ ∃y(L(y) ∧ H(x, y)) → ∃y(A(y) ∧ H(x, y)))]

    To prove, we shall follow a proof by contradiction. We shall include the negation of the conclusion for

    arguments. Since with just premise, deriving the conclusion is not possible, we have chosen this proof

    technique.

    Consider ∀x(L(x) → A(x)) ∧ ¬[∀x(L(x) ∧ ∃y(L(y) ∧ H(x, y)) → ∃y(A(y) ∧ H(x, y)))]

    We need to show that the above expression is unsatisfiable (False).

    ¬[∀x(L(x) ∧ ∃y(L(y) ∧ H(x, y)) → ∃y(A(y) ∧ H(x, y)))]

    ∃x¬((L(x) ∧ ∃y(L(y) ∧ H(x, y))) → ∃y(A(y) ∧ H(x, y)))

    ∃x((L(x) ∧ ∃y(L(y) ∧ H(x, y))) ∧ ¬(∃y(A(y) ∧ H(x, y))))

    E.I with respect to x,

    (L(a) ∧ ∃y(L(y) ∧ H(a, y))) ∧ ¬(∃y(A(y) ∧ H(a, y))), for some a

    (L(a) ∧ ∃y(L(y) ∧ H(a, y))) ∧ (∀y(¬A(y) ∧ ¬H(a, y)))

    E.I with respect to y,

    (L(a) ∧ (L(b) ∧ H(a, b))) ∧ (∀y(¬A(y) ∧ ¬H(a, y))), for some b

    U.I with respect to y,

    (L(a) ∧ (L(b) ∧ H(a, b)) ∧ (¬A(b) ∧ ¬H(a, b))), for any b

    Since P ∧ Q is P, drop L(a) from the above expression.

    (L(b) ∧ H(a, b)) ∧ (¬A(b) ∧ ¬H(a, b))), for any b

    Apply distribution

    (L(b) ∧ H(a, b) ∧ ¬A(b)) ∨ (L(b) ∧ H(a, b) ∧ ¬H(a, b))

    Note: P ∧ ¬P is false. P ∧ f alse is P. Therefore, the above expression is simplified to

    (L(b) ∧ H(a, b) ∧ ¬A(b))

    U.I of ∀x(L(x) → A(x)) gives L(b) → A(b). The contrapositive of this is ¬A(b) → ¬L(b). Replace

    ¬A(b) in the above expression with ¬L(b). Thus, we get,

    (L(b) ∧ H(a, b) ∧ ¬L(b)), this is again false.

    This shows that our assumption that the conclusion is false is wrong. Therefore, the conclusion follows

    from the premise.

    15

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